physics
posted by will .
A car is traveling 26 m/s when the driver sees a child standing in the road. He takes 0.8 s to react, then steps on the brakes and slows at 6.0 m/s2. How far does the car go before it stops?

In .8 seconds he goes 26 * .8 meters
then the problem is acceleration with initial speed Vo = 26 and acceleration = 6 m/s^2
time to deacceletate to 0:
v = Vo + a t
0 = 26  6 t
t = 26/6
distance = .8* 26 + 26 t  (1/2) 6 * t^2 
The distance travelled during the "reaction time" t1 is
X1 = V*t1 = 26 m/s * 0.8 s. = ?
The time required to decelerate to 0 is t2 = V/a = 4.33 s
The distance travelled while decelerating is the average velocity multipled by t2.
X2 = (V/2)(V/a) = V^2/(2a) = ?
Add X1 and X2 for the answer. 
Ah, good. I am off for a while :)